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4 months ago
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By Susam Pal on 27 May 2025
A field has exactly two ideals: the zero ideal, which contains only the additive identity, and the whole field itself. These are known as trivial ideals. Further, if a unital commutative ring, with distinct additive and multiplicative identities, has no ideals other than the trivial ones, then it must be a field. Together, these two facts reveal an interesting connection between trivial ideals and the structure of fields. In this article, we will explore why these facts are true.
Contents
Definition
A left ideal of a ring \( R \) is a subset \( I \subseteq R \) such that \( I \) is an additive subgroup of \( R, \) and for all \( r \in R \) and \( x \in I, \) we have \( r \cdot x \in I. \)
Similarly, a right ideal of a ring \( R \) is a subset \( I \subseteq R \) such that \( I \) is an additive subgroup of \( R, \) and for all \( r \in R \) and \( x \in I, \) we have \( x \cdot r \in I. \)
In a commutative ring \( R, \) every left ideal is also a right ideal, and vice versa. This is because for all \( r \in R \) and \( x \in I, \) we have \( r \cdot x = x \cdot r. \) Therefore, when working with commutative rings, we do not need to distinguish between left and right ideals. We simply refer to them as ideals. In this context, we also say that \( I \) is closed under multiplication by any element of the ring.
Let us look at an example before we proceed. The set of even integers \[ \langle 2 \rangle = \{ 2n : n \in \mathbb{Z} \} \] is an ideal of \( \mathbb{Z} \) generated by the integer \( 2. \) Indeed, if we multiply any integer by any even integer, the result is an even integer. In other words, if we multiply any element of \( \mathbb{Z} \) with an element of \( \langle 2 \rangle, \) the product belongs to \( \langle 2 \rangle. \)
Known Results
Familiarity with algebraic structures such as groups, rings, and fields is assumed. For the sake of brevity, we also assume the following standard results:
Proposition 1. Let \( R \) be a ring. Then, for all \( a \in R, \) we have \[ a \cdot 0 = 0 \cdot a = 0. \]
Proposition 2. Let \( R \) be a ring and let \( a \in R. \) Then \[ I_L = \{ r \cdot a : r \in R \}, \quad I_R = \{ a \cdot r : r \in R \} \] are, respectively, a left ideal and a right ideal of \( R. \) If \( R \) is commutative, then \( I_L = I_R, \) and we write \[ \langle a \rangle = \{ r \cdot a : r \in R \} \] and say that \( \langle a \rangle \) is an ideal of \( R \) generated by \( a. \)
Ideals of Fields
In this section, we show that a field \( K \) has only two ideals: \( \{ 0 \} \) and \( K \) itself.
Clearly \( \{ 0 \} \) is an ideal of \( K, \) as it satisfies the definition of an ideal. It is the trivial additive subgroup of \( K , \) and by Proposition 1, for all \( r \in K, \) we have \( r \cdot 0 = 0 \in \{ 0 \}. \)
Now \( K \) is also an ideal of itself. This follows directly. Since \( K \) is an additive group by the definition of a field, it is an additive subgroup of itself. Moreover, as a field, \( K \) is closed under multiplication, so for all \( r, x \in K \) we have \( r \cdot x \in K. \)
We will now show that \( \{ 0 \} \) and \( K \) are the only ideals of \( K. \) Let \( I \) be an ideal of \( K. \) There are two cases to consider: \( I = \{ 0 \} \) and \( I \ne \{ 0 \}. \) Suppose \( I \ne \{ 0 \}. \) Then there exists a non-zero element \( a \in I. \) Since \( a \ne 0 \) and \( K \) is a field, \( a \) has a multiplicative inverse \( a^{-1} \in K. \) Since \( I \) is closed under multiplication by any element of \( K , \) we have \[ a^{-1} \cdot a = 1 \in I. \] Now, let \( b \in K. \) Since \( 1 \in I, \) we get \[ b = b \cdot 1 \in I. \] Thus \( K \subseteq I, \) and since \( I \subseteq K \) by definition, we conclude \( I = K. \) Therefore the only ideals of \( K \) are \( \{ 0 \} \) and \( K \) itself.
Rings With Trivial Ideals
We now show that if \( R \) is a unital commutative ring with \( 1 \ne 0, \) and the only ideals of \( R \) are \( \{ 0 \} \) and \( R \) itself, then \( R \) must be a field. To do this, we first show that every non-zero element of \( R \) has a multiplicative inverse in \( R. \) Let \( a \in R \) with \( a \ne 0. \) We now show that there exists a multiplicative inverse \( a^{-1} \in R. \) Consider the ideal \[ \langle a \rangle = \{ r \cdot a : r \in R \}. \] Since \( a = 1 \cdot a \in \langle a \rangle, \) we have \( \langle a \rangle \ne \{ 0 \}. \) By assumption, the only ideals of \( R \) are \( \{ 0 \} \) and \( R, \) so it must be that \( \langle a \rangle = R. \) Since \( 1 \in R, \) we must have \[ s \cdot a = 1 \] for some \( s \in R. \) Thus \( a \) has a multiplicative inverse \( s \in R, \) and this holds for every non-zero \( a \in R. \)
The remaining properties of fields, namely, associativity and commutativity of addition and multiplication, the existence of distinct additive and multiplicative identities, the existence of additive inverses, and the distributivity of multiplication over addition, are inherited from the ring \( R. \) Therefore \( R \) is a field.
Conclusion
To summarise, any unital commutative ring with distinct additive and multiplicative identities that has only trivial ideals is a field, and every field has only trivial ideals.
